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The simple pendulum

The simple pendulum. θ. L. m. The simple pendulum. θ. L. m. mg. The simple pendulum. θ. L. m. mg sin θ. mg. The simple pendulum. θ. L. x. m. mg sin θ. mg. Some trig: sin θ = x L

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The simple pendulum

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  1. The simple pendulum θ L m

  2. The simple pendulum θ L m mg

  3. The simple pendulum θ L m mg sinθ mg

  4. The simple pendulum θ L x m mg sinθ mg

  5. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L x m mg sinθ mg

  6. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L + x m mg sinθ mg

  7. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ + x m mg sinθ mg

  8. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) + x m mg sinθ mg

  9. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L + x m mg sinθ mg

  10. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma + x m mg sinθ mg

  11. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

  12. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

  13. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L + x m mg sinθ mg

  14. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L Compare with SHM equation: a = - (2πf)2 x + x m mg sinθ mg

  15. and a = - g x L Compare with SHM equation: a = - (2πf)2 x θ L + x m mg sinθ mg

  16. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L θ L + x m mg sinθ mg

  17. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L θ L + x m mg sinθ mg

  18. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x m mg sinθ mg

  19. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x Discuss: effect of length, mass, gravity, angle of swing. m mg sinθ mg

  20. T = 2πL g

  21. T = 2πL g Put in the form: y = m x + c

  22. T = 2πL g Put in the form: y = m x + c T 2 = 4 π 2L + 0 g

  23. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m

  24. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mg

  25. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg r mg

  26. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg but r = L so mv2 = Ts - mg r L mg

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